Uniform Circular Motion

8 MCQs1 revision card9-step worked example

Source: NCERT KinematicsPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

The trap: aspirants see "constant speed" in uniform circular motion (UCM) and conclude there is zero acceleration. NEET exploits this confusion reliably — if you mark "acceleration = 0" because the speedometer doesn't change, you lose marks.

What UCM actually is. A particle moves along a circle of radius r at constant speed v. The speed (scalar) never changes. But velocity is a vector — it has both magnitude and direction. Because the direction of motion rotates continuously along the circle, the velocity vector changes every instant, which means there is a non-zero acceleration (NCERT Class 11 Physics Chapter 3, page 17).

Centripetal acceleration. This acceleration points radially inward — toward the centre of the circle — and has magnitude:

$$a_c = \frac{v^2}{r} = \omega^2 r$$

where ω = v/r is the angular speed. The word "centripetal" means "centre-seeking." There is no tangential acceleration component in UCM because the speed is constant — the entire acceleration is radial.

What stays constant, what doesn't.

Quantity	Constant in UCM?	Why
Speed	Yes	Definition of "uniform"
Kinetic energy (½mv²)	Yes	Depends only on speed
Velocity	No	Direction changes continuously
Acceleration magnitude	Yes	v²/r with constant v, r
Acceleration direction	No	Always points toward centre; rotates with the particle

The UCM-to-projectile bridge. A common NEET pattern gives you a particle in UCM (radius R, period T) and then asks: "if this particle is now launched vertically upward with the same speed, find the maximum height." The bridge step is computing v = 2πR/T from the circular motion data before applying projectile formulas. Do not set H = R — the radius and the projectile height are unrelated quantities.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Direct ApplicationPractice

A particle moves in a circle of radius 0.50 m at a constant speed of 4.0 m/s. What is the magnitude of its centripetal acceleration?

MCQ 2Easy RecallPYQ Pattern

In uniform circular motion at constant speed, which of the following quantities is NOT constant?

MCQ 3Concept TrapPractice

A body moves in a circle at constant speed. A student claims: 'Since the speed is constant, the acceleration is zero.' Which statement correctly identifies the error?

MCQ 4Easy RecallPractice

The centripetal acceleration of a particle in uniform circular motion is directed:

MCQ 5Direct ApplicationPractice

A particle in uniform circular motion has angular speed ω = 4.0 rad/s and moves on a circle of radius 0.25 m. What is the centripetal acceleration?

MCQ 6Easy RecallPractice

In uniform circular motion, the work done by the centripetal force over one complete revolution is:

MCQ 7CalculationPYQ Pattern

A particle moves in a circle of radius R with period T. If this particle is then launched vertically upward with the same speed it had in the circular motion, the maximum height reached is:

MCQ 8Direct ApplicationPractice

A stone tied to a string moves in a horizontal circle of radius 1.0 m at constant speed. If the magnitude of centripetal acceleration is 25 m/s², the speed of the stone is:

Quick recall before you leave

Worked Example

Pattern: UCM-to-projectile bridge (NEET pattern: projectile from circular motion, anchored to 2021 P3 Q49)

1
Given
- Radius R = 0.50 m - Period T = 0.20 s - g = 10 m/s² (exact, problem-defined)
2
Required
Maximum height H when launched vertically upward with speed v from UCM.
3
Concept
In UCM, the tangential speed is v = 2πR/T. When launched vertically, the particle undergoes uniformly decelerated motion under gravity. At maximum height, final velocity = 0. Use the kinematic relation: H = v²/(2g).
4
Formula
$$v = \frac{2\pi R}{T}, \qquad H = \frac{v^2}{2g}$$
5
Substitution
$$v = \frac{2\pi \times 0.50}{0.20} = \frac{\pi}{0.20} = 5\pi \text{ m/s}$$ $$H = \frac{(5\pi)^2}{2 \times 10} = \frac{25\pi^2}{20}$$
6
Calculation
$$H = \frac{25 \times 9.87}{20} = \frac{246.7}{20} = 12.3 \text{ m}$$ (Using π² ≈ 9.87.) Note on exact constants: g = 10 m/s² is stated as exact in the problem. The integers 2 in the denominator and the factor 2π are mathematical constants. These do not limit significant figures. The measured quantities (R = 0.50 m, T = 0.20 s) each have 2 significant figures, so the answer is reported to 2 significant figures: **H ≈ 12 m** (or 2π²R²/(gT²) = 12.3 m if left in exact form).
7
Final answer
$$H \approx 12 \text{ m}$$ Or in exact closed form: H = 2π²R²/(gT²).
8
Common trap
Setting H = R = 0.50 m. The circular radius and the projectile height are completely different quantities. The radius determines the UCM speed via v = 2πR/T; the height then follows from v²/(2g). Confusing geometric R with kinematic H gives an answer off by a factor of ~25 here.
9
Similar NEET-style question
A toy car moves in a horizontal circle of radius 0.80 m completing one revolution every 0.40 s. If the car is flicked vertically upward with its circular-motion speed, find the maximum height. (g = 10 m/s².) *Answer sketch:* v = 2π(0.80)/0.40 = 4π m/s; H = (4π)²/(20) = 16π²/20 ≈ 7.9 m. ---

Before solving, remember these

Law

Uniform circular motion — centripetal acceleration

An object moving in a circle of radius r with constant speed v has acceleration of magnitude a_c = v² / r directed toward the centre of the circle (centripetal). In terms of angular speed ω = v/r, a_c = ω² r. The acceleration changes direction continuously even though the speed is constant.

-- NCERT Class 11 Physics, Ch. 3, p. 17

Formulas

6 formulas — click to collapse

First kinematic equation (uniform acceleration)

v = v_{0} + a t

Final velocity equals initial velocity plus acceleration times the time elapsed, for motion under constant acceleration.

Symbol	Quantity	SI Unit
v	Final velocity	m/s
v0	Initial velocity	m/s
a	Constant (uniform) acceleration	m/s^2
t	Elapsed time	s

Valid when

Acceleration a is CONSTANT (uniform) in both magnitude and direction
All quantities measured in the same inertial reference frame
Motion is along a straight line; signs encode direction along chosen axis

Do NOT use when

Acceleration changes in magnitude or direction (use a(t) integration)
Motion is uniformly circular at constant speed (a is centripetal, not tangential)

Second kinematic equation (displacement under uniform acceleration)

x - x_{0} = v_{0} t + \frac{1}{2} a t^{2}

Displacement equals initial-velocity-times-time plus half of acceleration-times-time-squared. The (1/2) factor is the area of the triangle on the v-t graph.

Symbol	Quantity	SI Unit
x	Final position	m
x0	Initial position	m
v0	Initial velocity	m/s
a	Constant acceleration	m/s^2
t	Time elapsed	s

Valid when

Acceleration constant (magnitude and direction)
Sign convention consistent across x, v, a (one chosen positive direction)

Third kinematic equation (velocity-squared)

v^{2} = v_{0}^{2} + 2 a (x - x_{0})

Relates final velocity to initial velocity, displacement, and acceleration without using time. Most useful when t is unknown or unwanted.

Symbol	Quantity	SI Unit
v	Final velocity	m/s
v0	Initial velocity	m/s
a	Constant acceleration	m/s^2
x - x0	Displacement	m

Valid when

Constant acceleration
Use signed values for v, v0, a, and (x - x0) consistently

Do NOT use when

Time-dependent acceleration
Curvilinear motion where acceleration is not parallel to displacement

Projectile maximum height

H = \frac{v _{0}^{2} s i n ^{2} ( θ _{0} )}{2 g}

Maximum height attained by a projectile launched at speed v0 and angle theta0 above the horizontal, measured above the launch level.

Symbol	Quantity	SI Unit
H	Maximum height (above launch)	m
v0	Launch speed	m/s
theta0	Launch angle	rad/deg
g	Gravitational acceleration	m/s^2

Valid when

Air resistance neglected
Constant g over trajectory

Projectile horizontal range

R = \frac{v _{0}^{2} s i n ( 2 θ _{0} )}{g}

For a projectile launched from and returning to the same horizontal level with initial speed v0 at angle theta0 above the horizontal, the horizontal range R is given by this formula. R is maximised at theta0 = 45 deg.

Symbol	Quantity	SI Unit
R	Horizontal range	m
v0	Launch speed	m/s
theta0	Launch angle above horizontal	rad (or deg with sin in deg)
g	Gravitational acceleration	m/s^2

Valid when

Launch and landing are at the same vertical height
Air resistance neglected
g treated as constant over the trajectory

Do NOT use when

Launch and landing heights differ (use full kinematics)
Significant air drag (e.g. table-tennis ball, badminton shuttle)
Variation of g (ballistic trajectories spanning large altitude changes)

Centripetal acceleration in uniform circular motion

a_{c} = \frac{v ^{2}}{r} = ω^{2} r

An object moving in a circle of radius r at constant speed v has acceleration of magnitude v^2/r (or equivalently omega^2 * r) directed toward the centre. This is centripetal (radially inward), not tangential.

Symbol	Quantity	SI Unit
a_c	Centripetal acceleration	m/s^2
v	Tangential speed	m/s
r	Radius of circle	m
omega	Angular speed	rad/s

Valid when

Speed v is constant (uniform circular motion)
r and the centre are well-defined (instantaneous radius of curvature for general curved motion)

Do NOT use when

Non-uniform circular motion (then there is also a tangential acceleration component)

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

14 items — click to collapse

Trap

Free-fall: linear-in-time vs quadratic distance progression

Category: Similar Terms

Student answers 1:2:3:4 for distances in successive 1-second intervals (linear) instead of 1:3:5:7 (Galileo's odd numbers).

When it triggers

Question asks about ratios of distances traversed in successive 1-s intervals during free fall from rest.

How to avoid

Distance grows quadratically (y = ½ g t²); successive interval distances are y_n - y_{n-1} = ½ g (t_n² - t_{n-1}²) = ½ g (2n-1) seconds. The factor (2n-1) gives 1, 3, 5, 7, ...

Trap

Position-time graph slope: tan vs sin/cos

Category: Graph Interpretation

Student uses sin or cos of the angle the line makes with the time axis, instead of tan, to extract velocity.

When it triggers

Question gives an angle the x-t line makes with the t-axis (often 30°, 45°, 60°) and asks for velocity or its ratio.

How to avoid

Velocity = dx/dt = slope of x-t line = tan(angle), where the angle is measured from the time axis. Always tan, not sin or cos.

Trap

Implicit-function kinematics differentiation

Category: Overthinking

Student attempts to invert t(x) algebraically before differentiating, getting tangled in messy algebra; misses chain rule.

When it triggers

Question gives t as function of x (instead of x as function of t), e.g. t = x² + x.

How to avoid

Differentiate the given relation directly: dt/dx = (function of x). Then v = dx/dt = 1/(dt/dx). For acceleration use chain rule: a = dv/dt = (dv/dx)(dx/dt) = v dv/dx.

Trap

Initial velocity dropped from kinematic equation

Category: Sign Convention

Student treats a 'thrown vertically downward' problem as if the object were dropped (u = 0). The result is wrong by an additive u² term in v² = u² + 2gh. When the question explicitly states a launch speed, that speed is non-zero and CANNOT be ignored.

When it triggers

Question phrases: 'thrown vertically downward', 'projected with initial velocity', 'launched with speed u'. If u is given numerically, it MUST appear in the equation.

How to avoid

Always parse the launch verbal cue and write down u with its sign before reaching for v² = 2gh. Use the full v² = u² + 2gh (or u² - 2gh for upward motion).

Trap

Projectile max-height formula confused with range formula

Category: Similar Terms

Student plugs into v₀² sin(2θ)/g (range) when asked for maximum height, or vice versa. The two share v₀ and θ but have different sin-vs-sin² and 2g-vs-g terms.

When it triggers

Question mentions launch speed and angle and asks for max height (H) or range (R). Distractors include the wrong formula's answer.

How to avoid

Memorise BOTH formulas explicitly: H = v₀² sin² θ / (2g) (note sin²); R = v₀² sin(2θ) / g (note sin of doubled angle). Check by setting θ = 45°: max range, half max height.

Trap

Projectile angle: from horizontal vs from vertical

Category: Sign Convention

Student plugs angle θ into v cos θ when the question states 'angle with the vertical' (which makes the horizontal component v sin θ).

When it triggers

Question phrases like 'thrown at angle θ with the vertical direction' or 'with horizontal'.

How to avoid

Always identify reference axis explicitly. From horizontal: vx = v cos θ, vy = v sin θ. From vertical: vx = v sin θ, vy = v cos θ. The two are complementary (θ_h + θ_v = 90°).

Trap

Relative-velocity direction sign

Category: Sign Convention

Student fails to distinguish between same-direction and opposite-direction relative velocities, treating both as magnitudes.

When it triggers

Question describes two objects moving on the same line; observer somewhere between or alongside.

How to avoid

Relative velocity is a VECTOR. Same direction: v_rel = v_a - v_b (smaller magnitude). Opposite direction: v_rel = v_a + v_b (larger magnitude). Use sign convention consistently along chosen axis.

Trap

Linear speed-distance ratio under uniform deceleration

Category: Overthinking

Student assumes proportionality of speed to remaining distance under uniform deceleration. In fact, KE drops linearly with distance (v² is the linear quantity, not v): v² = u² - 2as. Speed-vs-distance is a sqrt-curve, not a line.

When it triggers

Question describes a body decelerating through stages with given speed at one stage; asks for distance to stop or speed at another stage.

How to avoid

Always work with v², not v, when uniform deceleration is in play. The work-energy theorem gives the same answer faster: ½ m v² = work done against constant force over distance.

Trap

UCM-to-projectile speed bridge

Category: Overthinking

Student uses the radius R as the projectile launch height or fails to compute the UCM speed from period.

When it triggers

Question describes a particle in UCM with given (R, T) then says 'now launched vertically up with same speed; find max height'.

How to avoid

Step 1: speed v = 2πR/T (from UCM). Step 2: max projectile height H = v²/(2g) = (2πR/T)² / (2g). Don't shortcut by setting H = R.

Trap

Uniform circular motion: velocity vs speed constancy

Category: Similar Terms

Student claims velocity is constant in uniform circular motion (it's not — direction changes).

When it triggers

Question asks 'in uniform circular motion at constant speed, which is also constant?'

How to avoid

In UCM: SPEED constant; KE constant. VELOCITY (vector) NOT constant. ACCELERATION (centripetal, magnitude v²/r) constant in MAGNITUDE but NOT in direction.

Mistake

Student treats average velocity as the arithmetic mean of two given speeds, or treats |average velocity| = average speed.

Root cause: concept gap

Correction

Average velocity = total displacement / total time (vector). Average speed = total path length / total time (scalar, always >= |average velocity|). For round-trip motion, average velocity is zero; average speed is not.

Wrong option pattern

Distractor offers (v1+v2)/2 instead of total-distance/total-time.

Mistake

Student applies v = v0 + a*t (or other constant-acceleration kinematic equations) to a problem where acceleration is not constant.

Root cause: formula misuse

Correction

The three kinematic equations require CONSTANT acceleration. For variable acceleration, use a = dv/dt and integrate, or use v dv = a dx for position-dependent acceleration. Verify constant-a before applying these formulas.

Wrong option pattern

Distractor uses constant-acceleration kinematic equations on a problem where the question explicitly says acceleration changes with time or position.

Mistake

Student forgets that projectile-motion formulas assume air resistance is negligible, or applies them to non-symmetric (different launch and landing height) trajectories without correction.

Root cause: concept gap

Correction

Standard range R = v0^2 sin(2*theta)/g and H = v0^2 sin^2(theta)/(2g) assume (i) launch and landing at the same height, (ii) negligible air drag, and (iii) constant g. For asymmetric trajectories, use the full kinematic decomposition along x and y.

Wrong option pattern

Distractor applies R = v0^2 sin(2*theta)/g to a projectile launched from a cliff.

Mistake

Student claims a body moving in a circle at constant speed has zero acceleration (because speed is constant).

Root cause: concept gap

Correction

Acceleration is the rate of change of VELOCITY (a vector), not speed. In uniform circular motion, the speed is constant but the velocity direction changes continuously, giving a centripetal acceleration of magnitude v^2/r toward the centre.

Wrong option pattern

Distractor option says 'a = 0 because speed is constant'.

Past Year Questions

10 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse

NEET 2025

In some appropriate units, time (t) and position (x) relation of a moving particle is given by t = x2 + x. The acceleration of the particle is 2 2

2− 2x+1 (x+2)3 2 2

3−

4+ (2x+1)3 (x+1)3

NTA Answer: Option 3(final)

NEET 2025

Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl is driving scooty with a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period T of the bus service and the speed (assumed constant) of the buses.

115 min, 120 km/h

29 min, 40 km/h

325 min, 100 km/h

410 min, 90 km/h

NTA Answer: Option 1(final)

NEET 2024Revised key

A particle moving with uniform speed in a circular path maintains:

1Constant velocity

2Constant acceleration

3Constant velocity but varying acceleration

4Varying velocity and varying acceleration

NTA Answer: Option 4(revised_final)

NEET 2023

A horizontal bridge is built across a river. A student standing on the bridge throws a small ball vertically upwards with a velocity 4 m s–1. The ball strikes the water surface after 4 s. The height of bridge above water surface is (Take g = 10 m s–2)

160 m

264 m

368 m

456 m

NTA Answer: Option 2(final)

NEET 2023

A bullet from a gun is fired on a rectangular wooden block with velocity u. When bullet travels 24 cm through u the block along its length horizontally, velocity of bullet becomes . Then it further penetrates into the block 3 in the same direction before coming to rest exactly at the other end of the block. The total length of the block is

124 cm

228 cm

330 cm

427 cm

NTA Answer: Option 4(final)

NEET 2022

The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second

11 : 1 : 1 : 1

21 : 2 : 3 : 4

31 : 4 : 9 : 16

41 : 3 : 5 : 7

NTA Answer: Option 4(final)

NEET 2022

The displacement-time graphs of two moving particles make angles of 30° and 45° with the x-axis as shown in the figure. The ratio of their respective velocity is

1- 6 - 1 : 3

23 : 1

31 : 1

41 : 2

NTA Answer: Option 1(final)

NEET 2022

A ball is projected with a velocity, 10 ms–1, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be

110 ms–1

2Zero

35 3 m s − 1

45 ms–1

NTA Answer: Option 3(final)

NEET 2021

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the 'θ' same speed at an angle to the horizontal, the maximum height attained by it equals 4R. The θ, angle of projection, is then given by : 1 2gT 2 2 θ=sin−1 

1π2  R  1  2  2 gT θ=cos−1

2  π2 R 1 π2  2 R θ=cos−1 

32 gT  1 π2  2 R θ=sin−1 

42 gT 

NTA Answer: Option 1(final)

NEET 2020

U U 20 m/s fl fl U – U 12. A ball is thrown vertically downward with a U 80 m/s fl U U – U ° velocity of 20 m/s from the top of a tower. It hits the ground after some time with a velocity of — (g=10 m/s2) 80 m/s. The height of the tower is : (g=10 m/s2)

1300 m

2360 m

3340 m

4320 m

NTA Answer: Option 1(final)

How NEET usually asks this

10 recurring patterns from past papers — click to collapse

For an object dropped from rest, distances traversed in successive equal time intervals are in the ratio 1:3:5:7:9:... (odd numbers). Derivable from y = ½ g t². Common shape: 'find ratio of distance covered in 1st, 2nd, 3rd, 4th seconds of free fall'.

Direct ApplicationEasy

Common distractors

uses arithmetic progression 1 2 3 4

Linear-time intuition

Time given as implicit function of position, e.g. t = x² + x. To find acceleration, differentiate twice: dt/dx = 2x + 1, so v = dx/dt = 1/(2x+1); a = dv/dt = -2v²·v = -2v³ (chain rule). Multi-step calculus.

Multi StepMedium

Common distractors

treats t x as explicit

Default to t-as-input thinking

A projectile (typically a bullet) penetrates a uniform medium with constant decelerating force; given initial speed and speed after a known distance, find the total stopping distance. Apply v² = u² + 2as to each segment, noting that 'a' is the same throughout. Common shape: bullet hits block at u, slows to u/k after distance d₁; how much further to stop?

Multi StepMedium

Common distractors

treats speed ratio as distance ratio

Linear thinking: 'speed went from u to u/3, so distance traversed should be 3× the original'

Two particles' position-time graphs are straight lines with given angles vs the time axis (e.g. 30° and 45°); find ratio of velocities. Velocity = slope = tan(angle). Common shape: ratio = tan(30°)/tan(45°) = 1/√3.

InterpretationEasy

Common distractors

uses sin or cos instead of tan

Trig confusion under time pressure

A particle in uniform circular motion (period T, radius R) is subsequently launched vertically upward with the SAME speed (v = 2πR/T); find max height. H = v²/(2g) = (2πR/T)²/(2g).

Multi StepMedium

Common distractors

uses radius as height

Surface confusion of geometric R and projectile H

Given launch speed v₀ and angle θ above horizontal, find maximum height. H = v₀² sin² θ / (2g). Common values plug in cleanly when sin θ ∈ {0.5, 0.707, 0.866}. Distractors test (i) using sin instead of sin², (ii) forgetting the factor of 2 in denominator, (iii) using g = 10 vs g = 9.8 inconsistently.

Direct ApplicationEasy

Common distractors

uses sin instead of sin squared

Confusing the height formula H = v²sin²θ/(2g) with the range 2v sin θ / g

Projectile launched at angle θ; find SPEED at highest point. The vertical component vanishes at the apex; horizontal component v₀ cos θ is conserved. Common gotcha: 'angle with the vertical' vs 'angle with the horizontal'.

Direct ApplicationEasy

Common distractors

uses sin instead of cos

Sign-of-angle-axis confusion

speed zero at top

True only for vertical motion (no horizontal v₀)

Buses leave both ends of a route every T minutes; observer travels between them at constant speed; given separate periods at which buses pass observer in same and opposite directions, find bus speed or T. Use relative-velocity equations: same direction T_same = L / (v_bus - v_obs), opposite T_opp = L / (v_bus + v_obs).

Multi StepMedium

Common distractors

ignores direction relative to observer

Treating bus-passing time as absolute

Particle in uniform circular motion at constant speed; identify which of speed / velocity / acceleration / kinetic energy is constant. Speed and KE constant; velocity (vector) and acceleration (centripetal) are NOT constant — direction changes.

InterpretationEasy

Common distractors

treats velocity as scalar

Conflates speed (scalar) with velocity (vector)

Object given a non-zero downward initial velocity from elevation; asked for the height fallen, time of impact, or final speed. Apply v² = u² + 2gh (or analogous) with proper signs. Common shape: a ball thrown vertically downward from a tower with initial speed u, hitting the ground at speed v; find the tower height. Distractors test (i) sign-of-u confusion, (ii) using v² = 2gh forgetting u², (iii) wrong g unit.

Multi StepEasy

Common distractors

drops initial velocity term

Student conflates 'thrown' with 'dropped' and uses v² = 2gh

Sources

NCERT refs: Class 11 Physics Chapter 3, p.17

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